when we square a real number it is impossible to get a negative real number.

**Introduction to
Complex Numbers**

Before introducing
complex numbers, let us try to answer the question “Whether there exists a real
number whose square is negative?” Let’s look at simple examples to get the
answer for it. Consider the equations 1 and 2.

This is because, when we square a real number it is impossible to
get a negative real number.

If equation 2 has solutions, then we must create an imaginary
number as a square root of −1.
This imaginary unit √−1
is denoted by *i *.The imaginary number *i *tells us that *i*^{2} = −1. We can use this fact
to find other powers of *i *.

We note that, for any
integer *n *, *i*^{n}* *has only four possible values: they
correspond to values of *n *when divided by 4 leave the remainders 0, 1,
2, and 3.That is when the integer *n *≤ −4 or *n *≥ 4 , using division algorithm, *n *can be written as *n *= 4*q *+ *k*, 0 ≤ *k *< 4, *k *and *q *are
integers and we write

*i*)^{n} = (*i*)^{4q+k} = (*i*)^{4q}(*i*)^{k} = ((*i*)^{4})^{q}(*i*)^{k} = (*i*)^{ q}(*i*)^{k} =(*i*)^{k}

**Example 2.1**

Simplify the following

(i) *i*^{7} (ii) *i*^{1729} (iii) *i*^{ −1924} + *i*^{2018} (iv) (v) *i* *i*^{2} *i*^{3} ….. *i*^{40}

**Solution**

(i) (*i*)^{7} = (*i*)^{4+3} = (*i*)^{3}_{ }= -*i*;

(ii) *i*^{1729} = *i*^{1729}*i*^{1} = *i*

(iii) (*i*)^{-1924} + (*i*)^{2018} =_{ }(*i*)^{-1924+0} + (*i*)^{2016+2} = (*i*)^{0} + (*i*)^{2} = 1-1 = 0

(iv) = (*i*^{1} + *i*^{2} +* i*^{3} +* i*^{4}) + (*i*^{5} +* i*^{6} +* i*^{7} +* i*^{8} ) + … + (*i*^{97} +* i*^{98} +* i*^{99} +* i*^{100} +) + *i*^{101} +* i*^{102}

= (*i*^{1} + *i*^{2} +* i*^{3} +* i*^{4}) + (*i*^{1} + *i*^{2} +* i*^{3} +* i*^{4}) + … + (*i*^{1} + *i*^{2} +* i*^{3} +* i*^{4}) + *i*^{1} +* i*^{2}

= {*i*+(-1)+(-*i*)+1} + {*i*+(-1)+(-*i*)+1} +
…. … + {*i*+(-1)+(-*i*)+1} + *i* + (-1)

= 0 + 0 + … 0 + *i* -1

= −1 + *i* (What is this number?)

(v) *i* *i*^{2} *i*^{3 … }*i*^{40 }=^{ }*i*^{1+2+3+…+40} = *i* ^{[40x41]/2} = *i*^{820} = *i*^{0} = 1

**Note**

(i) √[ab] = √a√b valid only if at least one of *a, b* is
non-negative.

For example, 6 = √36 = √[(-4)(-9)] = √(-4) √ (-9) = (2*i*) (3*i*)
= 6*i*^{2} = -6, a contradiction.

Since we have taken √[(-4)(-9)] = √(-4) √ (-9), we arrived at a
contradiction.

Therefore √[ab] = √a √b valid only if at least one of *a* *b*,
is non-negative.

(ii) For y ∈ **R** , y^{2} ≥ 0

* iy = yi.*

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12th Mathematics : UNIT 2 : Complex Numbers : Introduction to Complex Numbers |

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